I need post lab. please read them carefully. these are the wavelength values from my experiment.524nm for diethyl 2,2 cyanine iodide692nm diethyl 2,2 dicarbocyanine iodide604nm for diethyl 2,2 carbocynine iodide
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Chem 305 L – Spectra of Conjugated Dyes
UV-vis spectroscopy will be used to explore the electronic structure of a family of three
1,1- diethyl-2,2-cyanine iodide
1,1- diethyl-2,2-carbocyanine iodide
1,1- diethyl-2,2-dicarbocyanine iodide
A conjugated π-electron system is well represented by the particle-in-a-box (PIB) model. In this
model an electron is free to move within a box of fixed length where the potential inside the box
is zero. At the walls of the box, the potential is infinite. The potential difference causes the
electron to stay within the box.
As you have learned previously, the addition of energy can excite an electron from the highest
occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). The
HOMO energy level is found by taking the number of π -electrons in the conjugated chain, N,
and dividing by 2, a reflection of the Aufbau and Pauli Exclusion Principles. Each energy level in
our system can contain only two electrons. The next highest state is the LUMO. The assumption
we make in today’s experiment is that the wavelength of maximum absorbance (λ max)
represents the π -electron transition from the HOMO (N/2) to the LUMO (N/2 +1). The difference
in energy is given by
where Δn2 is [(N/2 +1)2 – (N/2)2], m is the mass of the electron, and L is the length of the box.
Since E =hv=hc/λ, we can obtain
Upon substituting in values for h (Planck’s constant), m (mass of electron), and c (speed of light)
and converting L and λ to nanometers, you can obtain an equation for L in terms of N and λ.
This derivation is part of your prelab.
It has been shown experimentally that for symmetric molecules, the PIB model needs to be
considered more carefully due to the effect of electron donating or withdrawing groups on either
side of the conjugated system. The potential at the walls is no longer infinite. As a result, the
length of the box appears to be shorter with electron donating groups and longer with electron
withdrawing groups. The correction parameter is α. If -1< α <1, then the PIB model is a good representation. The value of α allows us to quantify the extent to which the length of the box is extended or retracted. The parameter α should be consistent within a family of dyes, but has often times proven to be not. Calculation of another parameter will yield more consistent results while also allowing us to calculate the average bond length. L= (b x l) + γ (eq 3) In equation 3, b represents the number of carbon bonds in the chain (between nitrogens in our molecules), l is the average bond length along the chain in a series of dyes, and γ is a parameter that describes the change in the chain length due to electron wtihdrawing/donating groups. γ yields a consistent result within a series of dyes while a previous model that calculates α often did not. A plot of L (calculated from equation 2) vs b will give values of l from the slope and γ from the y-intercept. Since we have three dyes, we will have a graph with three points. Experiment: We will make stock solutions of 0.1mM dye with methanol. Instead of measuring out the tiny masses, we will simply use a few grains of dye in about 5-10mL of solvent. When we take the UV-vis, if the absorbance is beyond 2,we will dilute our samples until the absorbances are below 2. Scan the blank and samples from 350-750 nm. Pre lab: 1. Determine the structure of the 3 dyes. 2. Count the number of π -electrons in each molecule (along the carbon chain only). 3. What should we use for the blank? 4. Substitute values for h, m, and c and convert L and ƛ to nanometers in equation 2 (L2= (N+1)hλ,/8mc, to obtain an equation for L in terms of N and ƛ. Post-lab: The conjugated system in between nitrogen atoms of the cyanine dyes was analyzed. One nitrogen atom has a + charge and the other has a lone pair. The lone pair contributes to the extended π –orbital system. Consequently, the total number of π -electrons, designated the letter Nb is equal to the number of carbons in the conjugated system + 3 (or Nb= p+3). What is Nb for each of the molecules? Use the Nb value to calculate the L, length of the box, for each molecule. Use the equation derived in the pre-lab. Also, calculate the effective length of each carbon-carbon bond. Graph L vs. b to determine l and γ for the dyes. An alternate way to estimate the box length is to make the approximation that the conjugated bonds are essentially the same as 1.5 bonds in benzene, which are 0.139 nm. Estimate L by multiplying 0.139xNb for each of the molecules. Given the λmax’s for another family of dyes, do the same calculations as above for the following: 592 nm for 1,1'-diethyl-4,4'-cyanine iodide 708 nm for 1,1'-diethyl-4,4'-carbocyanine iodide 813 nm for 1,1'-diethyl-4,4'-dicarbocyanine iodide Show sample calculations, include your plots (properly labeled), and produce a table of your results. An example, which you should include in your plot, is given below for the molecule belonging to the 2nd set of dyes, 1,1'-diethyl-4,4'-tricarbocyanine iodide: L= 0.139Nb molecule b Nb λ (nm) L (N, λ ) c-c bond 1,1'-diethyl-4,4'tricarbocyanine iodide 14 16 929 2.19 0.136875 2.22 l (nm)* γ* * one value for series of dyes Compare the values of L (N, λ) and L = 0.139Nb. Comment on the agreement/pattern. Do they agree well or not? ... Our essay writing service fulfills every request with the highest level of urgency. attachment